What is the foci (y^2)/(16)-(x^2)/(20)=1 ?

The foci (y^2)/(16)-(x^2)/(20)=1 is (0,6),(0,-6)

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foci (y^2)/(16)-(x^2)/(20)=1

foci

\frac{y ^{2}}{16} - \frac{x ^{2}}{20} = 1

(0, 6), (0, - 6)

Solution steps

(h, k + c), (h, k - c)

Calculate Hyperbola properties

\frac{y ^{2}}{16} - \frac{x ^{2}}{20} = 1 :

Up-down Hyperbola with

(h, k) = (0, 0), a = 4, b = 25

(0, 0 + c), (0, 0 - c)

Compute

c :

c = 4^{2} + (25)^{2} : 6

(0, 0 + 6), (0, 0 - 6)

Refine

(0, 6), (0, - 6)

\frac{( y + 5 ) ^{2}}{64} - \frac{( x + 2 ) ^{2}}{36} = 1

\frac{x ^{2}}{1} - \frac{y ^{2}}{35} = 1

x^{2} - 36 y^{2} = 36

9 x^{2} - 9 y^{2} + 18 x + 36 y = 63

\frac{( y - 1 ) ^{2}}{\frac{1}{4}} - \frac{( x + 1 ) ^{2}}{\frac{1}{16}} = 1

What is the foci (y^2)/(16)-(x^2)/(20)=1 ?
The foci (y^2)/(16)-(x^2)/(20)=1 is (0,6),(0,-6)